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Answer:
The concentration of CH₃OH in equilibrium is [CH₃OH] = <em>2,8x10⁻¹ M</em>
Explanation:
For the equilibrium:
CO (g) + 2H₂(g) ⇄ CH₃OH(g) keq= 14,5
Thus:
14,5 = ![\frac{[CH_{3}OH]}{[CO][H_{2}]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DOH%5D%7D%7B%5BCO%5D%5BH_%7B2%7D%5D%5E2%7D)
In equilibrium, as [CO] is 0,15M and [H₂] is 0,36M:
14,5 = ![\frac{[CH_{3}OH]}{[0,15][0,36]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DOH%5D%7D%7B%5B0%2C15%5D%5B0%2C36%5D%5E2%7D)
Solving, the concentration of CH₃OH in equilibrium is:
<em>[CH₃OH] = 0,28M ≡ 2,8x10⁻¹ M</em>
I hope it helps!
<h3>Further explanation</h3>
Given
Reaction
H₂ (g) + I₂ (g) → 2HI(g)
Required
The equilibrium constant
Solution
The equilibrium constant is the value of the product in the equilibrium state of the substance in the right (product) divided by the substance in the left (reactant) with the exponents of each reaction coefficient
The equilibrium constant for reaction
pA + qB ⇒ mC + nD
![\large {\boxed {\bold {K ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7B%5Cbold%20%7BK%20~%20%3D%20~%20%5Cfrac%20%7B%5BC%5D%20%5E%20m%20%5BD%5D%20%5E%20n%7D%20%7B%5BA%5D%20%5E%20p%20%5BB%5D%20%5E%20q%7D%7D%7D%7D)
So for the above reaction :
![\tt K=\dfrac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=%5Ctt%20K%3D%5Cdfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
<u>Answer</u>:
Re-action of the working fluid may exceed the thrust force of the rocket.
<u>
</u><u>Explanation:</u>
A force that moves the rocket through the air present and then through space is known as thrust. This being generated by the 'propulsion system' of the rocket with newton’s third law of motion.
In the propulsion system, the engine works on the gas or liquid and that is further accelerated for working of the engine. This working fluid expelled in one direction whereas, the thrust force is applied in another direction may help in exceeding the weight of the rocket.